SATHEE: Area Question 40

Area Question 40

Question 40

If $A_{n}$ is the area bounded by the curve $y = (\tan x)^{n}$ and the lines $x = 0, y = 0$ and $x = \frac{π}{4}$ .

Then, prove that for $n > 2, A_{n} + A_{n + 2} = \frac{1}{n + 1}$ and deduce $\frac{1}{2 n + 2} < A_{n} < \frac{1}{2 n - 2}$ .

$(1996, 3 M)$

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Solution:

We have, $A_{n} = \int_{0}^{π / 4} (\tan x)^{n} d x$

Since, $0 < \tan x < 1$ , when $0 < x < π / 4$

We have, $0 < (\tan x)^{n + 1} < (\tan x)^{n}$ for each $n \in N$

$\Rightarrow \int_{0}^{π / 4} (\tan x)^{n + 1} d x < \int_{0}^{π / 4} (\tan x)^{n} d x$

$\Rightarrow A_{n + 1} < A_{n}$

Now, for $n > 2$

$\begin{aligned} A_{n} + A_{n + 2} & = \int_{0}^{π / 4} [(\tan x)^{n} + (\tan x)^{n + 2}] d x & = \int_{0}^{π / 4} (\tan x)^{n} (1 + \tan^{2} x) d x \end{aligned}$

$= \int_{0}^{π / 4} (\tan x)^{n} \sec^{2} x d x$

$= \frac{1}{(n + 1)} (\tan x)^{n + 1}$

$= \frac{1}{(n + 1)} (1 - 0) = \frac{1}{n + 1}$

Since, $A_{n + 2} < A_{n + 1} < A_{n}$ ,

then $A_{n} + A_{n + 2} < 2 A_{n}$

$\Rightarrow \frac{1}{n + 1} < 2 A_{n}$

$\Rightarrow \frac{1}{2 n + 2} < A_{n}$

Also, for $n > 2 A_{n} + A_{n} < A_{n} + A_{n - 2} = \frac{1}{n - 1}$

$\begin{array}{ll} \Rightarrow & 2 A_{n} < \frac{1}{n - 1} \Rightarrow & A_{n} < \frac{1}{2 n - 2} \end{array}$

From Eqs. (i) and (ii), $\frac{1}{2 n + 2} < A_{n} < \frac{1}{2 n - 2}$

Area Question 40

Question 40

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Area Question 40

Question 40

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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